How to show that e.g. $\cos(z)$ is analytic using Cauchy-Riemann differential equations [$u_x(x,y)=v_y(x,y)$ and $u_y(x,y)=-v_x(x,y)$]? Do all analytic functions satisfy Cauchy-Riemann differential equations (CRDE)? What is the relationship between analyticity of complex functions and Cauchy-Riemann differential equations? I know that holomorphic (analytic?) functions satisfy CRDE, but are functions that satisfy CRDE always analytic (holomorphic)?
[Math] How to show that e.g. $\cos(z)$ is analytic using Cauchy- Riemann differential equations
complex-analysis
Related Solutions
There is a difference between real differentiation and complex differentiation. The function $f(x+iy)=x^2$ is real differentiable as a function of $x$ (with derivative $2x$) but, as you showed from applying the Cauchy-Riemann equations, is not complex differentiable as a function of $z=x+iy$.
The question is: why is a function that is real differentiable not necessarily complex differentiable (as you've shown)? Recall that the derivative of a function $f(x)$ is defined as $$ f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}. $$ If $f$ is function of a real number then $h$ has to be real. Roughly speaking, the real number $h$ can only approach the limit 0 from two directions. Either you let $h$ be positive and make it smaller and smaller, or you let $h$ be negative and make it larger and larger, so approaching 0 in two ways. For the limit to exist these two ways of approaching 0 have to give the same result. This is why the function $f(x)=|x|$ is not differentiable at $x=0$: the result depends on whether you start with $h$ negative (giving "$f'(x)=-1$") or $h$ positive (giving "$f'(x)=+1$").
On the other hand, if $f$ is a function of a complex variable then $h$ is also complex. If $h$ is complex then it can approach 0 in an infinite number of ways. Think of the two dimensional complex plane: you can travel towards the origin $z=0$ though any of the infinite number of lines that pass through $z=0$. In particular, we could let $h$ approach 0 along the imaginary axis by setting $h=it$ for a real number $t$.
Let's try this out directly and show that your function $f(x+iy)=x^2+0i$ is not complex differentiable at $z=x+iy=1$. If we let $h$ be real then we get $$ f'(1) = \lim_{h\rightarrow 0}\frac{f(1+h)-f(1)}{h}\\ f'(1) =\lim_{h\rightarrow 0}\frac{(1+h)^2-(1)^2}{h}\\ f'(1) =\lim_{h\rightarrow 0}\frac{1+2h+h^2-(1)^2}{h}\\ f'(1) =\lim_{h\rightarrow 0}(2+h) = 2 $$ On the other hand, if we let $h$ be purely imaginary by setting $h=it$ for a real $t$ we get $$ f'(1) = \lim_{h\rightarrow 0}\frac{f(1+it)-f(1)}{h}\\ f'(1) =\lim_{h\rightarrow 0}\frac{(1)^2+0t-(1)^2}{h}\\ f'(1) =\lim_{h\rightarrow 0}\frac{0}{h}=0 $$ Thus when we allow $h$ to be complex it is possible for the limit defining the derivative to take on different values depending on the way in which you approach 0, and hence the limit doesn't exist.
Here's the Taylor expansion approach. First, we need a linear algebra fact:
Any $\mathbb R$-linear map $T:\mathbb C\to\mathbb C$ can be written as $Tz=\alpha z+\beta\bar z$ for some (unique) $\alpha,\beta\in \mathbb C$.
Uniqueness should be clear, and existence follows by first writing $T(x+iy)=\gamma x+\delta y$ with complex $\gamma,\delta$ and then replacing $x=(z+\bar z)/2$, $y=(z-\bar z)/(2i)$.
This linear algebra fact is worth remembering, as it simplifies various computations in complex analysis.
Back to the problem. The first-order real Taylor expansion of $f$ at a point $z\in\mathbb C$ takes the form $$ f(z+h)=f(z)+\alpha h+\beta \bar h+o(|h|) $$ and we want to show that $\beta=0$ here. The equations you are given say that $r f_r + i f_\theta = 0$. Write $h=r e^{i\theta}$, and take the derivatives of $\alpha h+\beta \bar h$: $$ \begin{align} f_r &= \alpha e^{i\theta} + \beta e^{-i\theta} \\ f_\theta & = \alpha ir e^{i\theta} - \beta i r e^{-i\theta} \\ rf_r+f_\theta & = 2\beta r e^{-i\theta} \end{align} $$ Thus $\beta=0$ as desired.
Best Answer
Start by rewriting: if $z$ is complex, then let $z=x+iy$. Then we have the function $\cos(x+iy)$. Now you can expand that with the rule $\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$. (Because you'll be left with bits like e.g. $\cos(iy)$, you'll may want to replace these with hyperbolic functions with e.g. $\cos(ip)=\cosh(p)$ and a similar relationship for $\sin$.) You'll be left with some complex function which we'll call $u+vi$ - i.e. let $u$ be the real part and $v$ the imaginary part. It is these $u$ and $v$ you are differentiating in the Cauchy-Riemann equations.